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Vlad1618 [11]
1 year ago
6

An engineer designed a valve that will regulate water pressure on an automobile engine. The engineer designed the valve such tha

t it would produce a mean pressure of 5.4 pounds/square inch. It is believed that the valve performs above the specifications. The valve was tested on 24 engines and the mean pressure was 5.7 pounds/square inch with a standard deviation of 1.0. A level of significance of 0.05 will be used. Assume the population distribution is approximately normal. Determine the decision rule for rejecting the null hypothesis. Round your answer to three decimal places.
Mathematics
1 answer:
KiRa [710]1 year ago
6 0

The p-value from the hypothesis test is 0.142 i.e., greater than the given significance level of 0.05. So, the null hypothesis is not rejected. The z-score for the given sample is 1.471.

<h3>What is the decision rule for the p-value approach to hypothesis testing?</h3>

The decision rule based on p-value states,

  • If p > α (significance level), then the null hypothesis is not rejected
  • If p < α (significance level), then the null hypothesis is rejected in favor of the alternative hypothesis.

<h3>Calculation:</h3>

Since it is given that the valve would produce a mean pressure of 5.4 pounds/square inch. I.e., μ = 5.4 p/si

So, Defining the hypothesis:

Null hypothesis H0: μ = 5.4

Alternative hypothesis Ha: μ ≠ 5.4

It is given that,

The valve was tested on 24 engines. I.e., Sample size n = 24

The sample mean X = 5.7

Standard deviation σ = 1.0 and

The significance level = 0.05

Since the population distribution is approximately normal,

the test statistic is calculated as follows:

z = (X - μ)/(σ/\sqrt{n})

On substituting the value,

z = (5.7 - 5.4)/(1.0/\sqrt{24})

  = (0.3)/0.204

  = 1.471

Fron this z-score, the p-value is calculated as 0.142.

Since, the value of p > 0.05 (significance level), the null hypothesis is not rejected.

Learn more about hypothesis testing here:

brainly.com/question/22078281

#SPJ4

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Olenka [21]

Answer:

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Step-by-step explanation:

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8 0
3 years ago
2. When a large truckload of mangoes arrives at a packing plant, a random sample of 150 is selected and examined for
kirza4 [7]

a) The 90% confidence interval of the percentage of all mangoes on the truck that fail to meet the standards is: (7.55%, 12.45%).

b) The margin of error is: 2.45%.

c) The 90% confidence is the level of confidence that the true population percentage is in the interval.

d) The needed sample size is: 271.

<h3>What is a confidence interval of proportions?</h3>

A confidence interval of proportions has the bounds given by the rule presented as follows:

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

The margin of error is given by:

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

The variables are listed as follows:

  • \pi is the sample proportion, which is also the estimate of the parameter.
  • z is the critical value.
  • n is the sample size.

The confidence level is of 90%, hence the critical value z is the value of Z that has a p-value of \frac{1+0.90}{2} = 0.95, so the critical value is z = 1.645.

The sample size and the estimate are given as follows:

n = 150, \pi = \frac{15}{150} = 0.1

The margin of error is of:

M = z\sqrt{\frac{0.1(0.9)}{150}} = 0.0245 = 2.45\%

The interval is given by the estimate plus/minus the margin of error, hence:

  • The lower bound is: 10 - 2.45 = 7.55%.
  • The upper bound is: 10 + 2.45 = 12.45%.

For a margin of error of 3% = 0.03, the needed sample size is obtained as follows:

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

0.03 = 1.645\sqrt{\frac{0.1(0.9)}{n}}

0.03\sqrt{n} = 1.645\sqrt{0.1(0.9)}

\sqrt{n} = \frac{1.645\sqrt{0.1(0.9)}}{0.03}

(\sqrt{n}})^2 = \left(\frac{1.645\sqrt{0.1(0.9)}}{0.03}\right)^2

n = 271 (rounded up).

More can be learned about the z-distribution at brainly.com/question/25890103

#SPJ1

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